Question

A rod of length L has non-uniformly distributed
mass along its length. For its mass per unit length
mo
varying with distance x from one end as
L?
L+x.
Find the position of centre of mass of this system.
O
L/4
L/2
O
3L/9
O
5L/9​

Answer 1

Answer:

ANSWER

X  

CM

​  

=  

∫  

0

L

​  

dm

∫  

0

L

​  

xdm

​  

=  

∫  

0

L

​  

 

L  

2

 

m  

0

​  

 

​  

(L+x)dx

∫  

0

L

​  

x  

L  

2

 

m  

0

​  

 

​  

(L+x)dx

​  

 

=  

∫  

0

L

​  

(L+x)dx

∫  

0

L

​  

x(L+x)dx

​  

=  

∫  

0

L

​  

(L+x)dx

∫  

0

L

​  

(xL+x  

2

)dx

​  

 

=  

L(x)  

0

L

​  

(  

2

x  

2

 

​  

)  

0

L

​  

 

(  

2

x  

2

 

​  

L)  

0

L

​  

+(  

3

x  

3

 

​  

)  

0

L

​  

 

​  

=  

L  

2

+  

2

L  

2

 

​  

 

2

L  

3

 

​  

+  

3

L  

3

 

​  

 

​  

 

=  

2

3L  

2

 

​  

 

6

3L  

3

+2L  

3

 

​  

 

​  

=  

6

5L  

3

 

​  

×  

3L  

2

 

2

​  

=  

9

5

​  

L

Explanation:

Answer 2

Appropriate Question :-

A rod of length L has non-uniformly distributed mass along its length. For its mass per unit length varying with distance x from one end as $\displaystyle \sf \frac{ m_{0}}{L^{2} }(L + x)$, Find the position of centre of mass of this system when x = 0 to x = L

Solution :-

$\qquad \displaystyle \sf \frac{dm}{dx} = \: \frac{m_{0}}{ {L}^{2} } \: (L + x)$

$\qquad \displaystyle \sf dm = \: \frac{m_{0}}{ {L}^{2} } \: (L + x)dx$

$\qquad \sf x_{cm} = \frac{ \: \displaystyle \sf∫ _{0}^{L}x \: \cancel{\frac{m_{0}}{ {L}^{2}} } \: (L + x)dx}{ \displaystyle \sf∫ _{0}^{L}\: \cancel{\frac{m_{0}}{ {L}^{2}} } \: (L + x)dx}$

$\qquad \sf = \frac{ \: \displaystyle \sf∫ _{0}^{L} {x}^{'} L dx +{{ ∫ _{0}^{L} {x}^{2} } }dx}{ \displaystyle \sf∫ _{0}^{L}\:L dx +∫ _{0}^{L} {x}^{'} L dx }$

$\qquad \sf = \frac{ \: \displaystyle \sf L + \frac{ {x}^{2} }{2} ∫ _{0}^{L}+ \: \frac{ {x}^{2} }{3} {{ ∫ _{0}^{L} } }}{ \displaystyle \sf L \: x∫ _{0}^{L}+ \: \frac{ {x}^{2} }{2} ∫ _{0}^{L} }$

$\qquad \sf = \frac{ \: \displaystyle \sf \frac{ {L}^{3} }{2} +\frac{ {L}^{3} }{3} }{ \displaystyle \sf L + \frac{ {L}^{2} }{2} }$

$\qquad \sf = \frac{ \: \displaystyle \sf {L}^{3} \bigg( \frac{ 1 }{2} +\frac{ 1 }{3} \bigg )}{ \displaystyle \sf {L}^{2} \bigg(1 + \frac{ 1 }{2} \bigg) }$

$\qquad \sf = \frac{ \: \displaystyle \sf {L} \bigg( \frac{ 5 }{6} \bigg)}{ \displaystyle \sf \frac{ 3 }{2} }$

$\displaystyle\qquad {\bf = \frac{ 5 L}{9 }}$

$\:$